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Nataliya [291]
2 years ago
7

You whirl a stone in a horizontal circle in such a way that the stone is in uniform circular motion. Which of the following is t

rue for this situation? (Select all that apply.)
a. The direction of the stone's velocity changes as it moves around the circle.
b. The magnitude of the stone's velocity does not change.
c. The stone's speed is constant.
d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.
e. The stone is moving with constant velocity.
Physics
1 answer:
horsena [70]2 years ago
4 0

a. The direction of the stone's velocity changes as it moves around the circle.

b. The magnitude of the stone's velocity does not change.

d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.

Above given are true for the given situation.

<u>Answer:</u> Option A, B and D

<u>Explanation:</u>

Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.

The velocity, a vector quantity is constant in a uniform circle motion speed is constant as its direction continues to change. Centripetal force works inward toward the core to counterbalance the centrifugal force from the center moving outward.

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Water has the ability to dissolve polar covalent molecules like sucrose by
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Explanation:

Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their

dipole - dipole attractions and hence hydrogen bonds are formed.

However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.

5 0
2 years ago
Find expressions for the force needed to bring an object of mass m from rest to speed v in time t. express your answer in terms
VARVARA [1.3K]
Good morning.

We have that:

\mathsf{V = a\cdot t} , since we have rest in the inicial time.

The acceleration can be found with Newton's Law:

\mathsf{F = m\cdot a\iff a = \dfrac{F}{m}}

Now we put the acceleratin in the velocity equation:

\mathsf{V = \dfrac{F}{m} \cdot t}

We want the force, so, let's isolate F:

\mathsf{V\cdot m = F\cdot t}\\ \\ \\ \boxed{\mathsf{F = \dfrac{V\cdot m}{t}}}

3 0
2 years ago
I Need answer ASAP
anzhelika [568]

Answer: Convection and conduction

Tell me that I got it right??

Explanation

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3 0
3 years ago
Read 2 more answers
Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be
avanturin [10]

Answer:

The mass rate of the cooling water required is: 1'072988.5\frac{kg}{h}

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

{m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}

Where w refers to the cooling water and s to the steam flow. Reorganizing,

m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}

Write the difference of enthalpy for water as Cp (Tout-Tin):

m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, {h_{s}}^{out}=251.40\frac{kJ}{kg} and {h_{s}}^{in} can be calculated as:

{h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}.

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 \frac{kJ}{kgK}. If the average temperature is actually different, it won't mean a considerable mistake. Also we know that {T_{w}}^{out}-{T_{w}}^{in}\leq 10, so let's work with the limit case, which is {T_{w}}^{out}-{T_{w}}^{in}=10 to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}

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5 0
3 years ago
A car is moving in uniform circular motion. If the car's speed were to double to keep the car moving with the same radius, the a
Anestetic [448]

Answer:

Increase by a factor of 4.

Explanation:

The acceleration of a car moving with speed v in a circle of radius R is given by:

a=\frac{v^2}{R}.

Now if we double the speed v in the equation above, it becomes 2v. Thus:

a=\frac{v^2}{R}\:\: {\rightarrow}\:\: a_n=\frac{(2v)^2}{R}=4\frac{v^2}{R}=4a.

Therefore the acceleration is increased by a factor of 4.

7 0
3 years ago
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